How do you find #int (x+4)/(x^2+2x+5) dx# using partial fractions?

Question

How do you find #int (x+4)/(x^2+2x+5)  dx# using partial fractions?

Christopher Allers · Accepted Answer

#int (x+4)/(x^2+2x+5) dx = 1/2 ln (x^2+2x+5) + 3/2 tan^(-1)((x+1)/2) + C#

Since we are requested to use partial fractions to solve this, I guess the hope is that we can simplify the integrand.

However, the quadratic #x^2+2x+5# only has Complex zeros, so if we do want to simplify the expression, we end up with Complex coefficients.

They say fools rush in where angels fear to tread, so here I go...

#x^2+2x+5 = (x+1)^2+2^2#

#color(white)(x^2+2x+5) = (x+1)^2-(2i)^2#

#color(white)(x^2+2x+5) = ((x+1)-2i)((x+1)+2i)#

#color(white)(x^2+2x+5) = (x+1-2i)(x+1+2i)#

So:

#(x+4)/(x^2+2x+5) = A/(x+1-2i)+B/(x+1+2i)#

Using Heaviside's cover up method, we find:

#A = (color(blue)(-1+2i)+4)/(color(blue)(-1+2i)+1+2i) = (3+2i)/(4i)= 1/2-3/4i#

#B = bar(A) = 1/2+3/4i#

So:

#int (x+4)/(x^2+2x+5) dx#

#= int (1/2-3/4i) * 1/(x+1-2i) + (1/2+3/4i) * 1/(x+1+2i) dx#

#= (1/2-3/4i) ln (x+1-2i) + (1/2+3/4i) ln (x+1+2i) + C#

#= 1/2(ln (x+1-2i) + ln (x+1+2i))+3/4 i (ln(x+1+2i)-ln(x+1-2i)) + C#

#= 1/2 ln (x^2+2x+5)+3/4 i (ln(x+1+2i)-ln(x+1-2i)) + C#

Use:

#tan^(-1)(z) = 1/2i(ln(1-iz)-ln(1+iz))#

So:

#3/4 i (ln(x+1+2i)-ln(x+1-2i))#

#= 3/4 i (ln(1+i(2/(x+1))) - ln(1-i(2/(x+1))))#

#= -3/2 1/2 i (ln(1-i(2/(x+1))) - ln(1+i(2/(x+1))))#

#= -3/2 tan^(-1)(2/(x+1))#

#= 3/2 tan^(-1)((x+1)/2) + (3pi)/4#

Hence:

#int (x+4)/(x^2+2x+5) dx = 1/2 ln (x^2+2x+5) + 3/2 tan^(-1)((x+1)/2) + C#

Scarlett Allison · Answer

undefined To find the integral $\int \frac{x + 4}{x^2 + 2x + 5} \, dx$ using partial fractions, you first decompose the fraction into simpler fractions. Then you solve for the coefficients of the partial fractions by equating the original expression with the decomposition. After finding the coefficients, you integrate each partial fraction separately.